tại x= bao nhiêu ạ

9 lần

tick tui tui nói cách giải cho

A=\(2x^2+5x+7=2\left(x^2+2.\dfrac{5}{4}x+\dfrac{25}{16}+\dfrac{31}{16}\right)=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{31}{8}\)

ta thấy \(2\left(x+\dfrac{5}{4}\right)^2\ge0=>2\left(x+\dfrac{5}{4}\right)^2+\dfrac{31}{8}\ge\dfrac{31}{8}\)

dấu ''='' xảy ra khi \(x+\dfrac{5}{4}=0< =>x=-\dfrac{5}{4}\)

vậy min A=31/8<=>x=-5/4

chúc bạn học tốt ^^

90360+65278=155638

tick mình nha mọi người,lạnh quá...hừu hừ.....

\(\left(x^2+x\right)^2+4x^2+4x+3=0\)

<=>\(\left(x^2+x\right)^2+4\left(x^2+x\right)+3=0\)

đặt \(x^2+x=y\)ta được:

\(y^2+4y+3=0< =>y^2+y+3y+3=0< =>y\left(y+1\right)+3\left(y+1\right)=0< =>\left(y+1\right)\left(y+3\right)=0=>\left[{}\begin{matrix}y+1=0\\y+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}y=-1\left(1\right)\\y=-3\left(2\right)\end{matrix}\right.\)

giải (1) :

\(x^2+x=-1< =>x^2+x+1=0< =>x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}=0\)

=>\(\left(x+\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\left(vônghieemj\right)\)

giải 2:

\(x^2+x=-3< =>x^2+x+3=0< =>x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{11}{4}=0\)

<=>\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0< =>\left(x+\dfrac{1}{2}\right)^2=-\dfrac{11}{4}\left(vônghiemej\right)\)

vậy.................

chúc bạn hcoj tốt ^^

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